Solving Homogeneous Differential Equation MCQ Quiz in मल्याळम - Objective Question with Answer for Solving Homogeneous Differential Equation - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Calculation

Put   ⇒ 

Integrating we get;

⇒ 

⇒ 

Putting (1,1)

⇒ 

⇒ 

hence its is a circle of radius 1

Hence option 2 is correct

Calculation

Given

f(1) = 1,  f(a) = 0 

f(x) = 

⇒ 

⇒  

⇒ 

⇒ 

⇒ 

Put 

⇒ 

⇒ 

Integrating both side 

⇒ e^v (x + c) + 1 + v = 0 

For f(1) = 1 ⇒ c = 

⇒ 

⇒ 

For 

⇒ ae = 2 

Calculation

Given

Let x = X + h, y = Y + k

⇒ 

Let Y = vX

⇒ 

⇒ 

⇒ 

⇒ 

As curve is passing through (2, 1) ⇒ C = 0

⇒ 

∴ α = 1 and β = 2

⇒ 5β + α = 11

Calculation

Given

⇒ 

Put 

⇒ 

⇒ 

⇒ 

⇒ 

⇒ 

⇒ cy = 

Put x = e , y = 1 ⇒ c = 1

Given, xy2dy - (x3 + y3)dx = 0 

⇒ xy2dy = (x3 + y3)dx

⇒ 

⇒ 

Put y = vx → 

⇒ 

Given, x dy - y dx = 

⇒ x dy = y dx +  

⇒ 

⇒ 

Put y = vx →  

⇒ 

⇒ 

⇒ 

Integrating both sides, 

⇒ d

⇒ 

⇒ 

⇒ 

∴ The correct answer is option (2).

Given, xy2dy - (x3 + y3)dx = 0 

⇒ xy2dy = (x3 + y3)dx

⇒ 

⇒ 

Put y = vx → 

⇒ 

Calculation:

We wish to solve the differential equation

Since both numerator and denominator are homogeneous of degree 2, set

Substituting gives

Separate variables:

Notice the partial‐fraction expansion

Hence

So

Since \), we rewrite in terms of x+y:

Apply the initial condition y(1)=0 to get (C=0). Therefore the solution is

Hence, the correct answer is Option 3. 

Calculation

Given equation:

Divide by x:

Let y = vx, then

Substitute in the equation:

⇒ 

⇒ 

⇒ 

Integrate both sides:

⇒ 

⇒ 

Remove the logarithms:

⇒ 

Substitute v = y/x:

⇒ 

∴ The general solution is .

Hence option 2 is correct

Concept:

If  is a homogeneous differential equation, put y = vx,

Then substitute .

Calculation:

Given 

⇒ 

⇒ (xy cos + y² sin )dx = (xy sin - x² cos )dy

⇒ 

Dividing the numerator and denominator of R.H.S by x².

⇒ 

Put y = vx

⇒ 

∴ Rewriting the equation, we get:

v + x = 

⇒  x = 

⇒ 

Integrating both sides, we get:

⇒ 

⇒ 

⇒ log |sec v| - log |v| = 2 log |x| + log k.

⇒ log  = log |x²| + log k.

⇒  log  - log |x2| = log k.

⇒ log  = log k.

⇒ log  = log k

⇒ 

⇒ sec  = kxy

⇒